Discussion in 'Homework Help' started by saki, Jun 26, 2008.

1. ### saki Thread Starter New Member

Jun 26, 2008
5
0
Using the method of determinants, calculate the values of the circulating currents in the circuit shown in fig.8 below:
Fig.8

-j

1+j3 V

5V

-j2

+j3

2. ### saki Thread Starter New Member

Jun 26, 2008
5
0
Check the attachment for the diagram

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May 16, 2005
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6
4. ### saki Thread Starter New Member

Jun 26, 2008
5
0
My attempt on the 8the question is this way:

Let the current in loop having 5V be I1 and that in the other loop be I2.
So we get,

I1 + 3I1 + 3I2 + j3I1 –jI1 –jI2 -5 = 0
(4+ j3 –j) I1 – jI2 – 5 = 0
4+j2) I1 –jI2 = 5 …1

Similarly, in Loop 2 we get,

3I2 + 3I2 + 3I1 –j2I2 –JI2 – 1 – j3 = 0
3I1 + (6 –j2 –j)I2 –(1+j3) = 0
3I1 + (6-j3)I2 = 1+j3 ……2

Taking both eq 1 and 2 into consideration we get,

[ (4+j2) -j] [ I1 ] = 5
3 (6-j3) I2 1+j3

5. ### saki Thread Starter New Member

Jun 26, 2008
5
0
Check the .doc attached here...

I gave my attempt at question 8.

For question 9 I know this is transformer issue and

kVA of a single phase transformer = V x A
and Z = V/I

but I was told that I should consider the frequency and other factors into place so what do i do abt that particular point?

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6. ### JoeJester AAC Fanatic!

Apr 26, 2005
3,373
1,159
Your figure 9 mentions the frequency.

7. ### saki Thread Starter New Member

Jun 26, 2008
5
0
can you please guide me on how should i be solving it.

I just need guidance on how to solve it.

8. ### Ratch New Member

Mar 20, 2007
1,068
3
saki,

Here is the way I think prob 8 should be set up. You appear able to solve the two simultaneous linear equations. Ratch

Code ( (Unknown Language)):
1.   (4+j2)I1         -(3-j)I2       = 5
2.   -(3-j)I1         (6-j3)i2       = -(1+j3)

9. ### Ratch New Member

Mar 20, 2007
1,068
3
saki,

The inductive reactance of a 0.1H coil at 50Hz is 2*3.14*50*0.1 = 31.42 ohms

The mutual inductance of the transformer is M = 0.9*sqrt(L1*L2) = 0.9*0.1=0.09H, and the inductive reactance is 2*3.15*50*0.09 = 28.27

Assume a load across A-B of ZL

Setting up the two node equations

Code ( (Unknown Language)):
1.     (10+j31.42)I1     -(j28.27)I2    =  5
2.    -(j28.27)I1        (10+j31.42+ZL) =  0
Solving, we get

I1 = [(10+j31.42+ZL)*5]/[(10+j31.42)(10+j31.42+ZL)+28.27^2]

For open circuit ZL = infinite so I1 = 5/(10+j31.42) = .05-j0.14

For short circuit ZL = 0 so I1 = 5/[(10+j31.42)+(28.27^2/(10+j31.42)] = 0.23-j0.11

I will let you substitute 10-j10 for ZL and solve for I1

Ratch